Given that $X$ is a random variable uniformly distributed on the interval $[7, 16]$, the probability density function (PDF) for a uniform distribution on $[a, b]$ is:

$f(x) = \frac{1}{b - a}, \quad \text{for} \quad a \leq x \leq b$

For the interval $[7, 16]$, we have $a = 7$ and $b = 16$.

Steps to find the standard deviation:

1. Mean (Expected Value): The mean $\mu$ of a uniform distribution on $[a, b]$ is given by: $\mu = \frac{a + b}{2}$ Substituting $a = 7$ and $b = 16$: $\mu = \frac{7 + 16}{2} = \frac{23}{2} = 11.5$

2. Variance: The variance $\sigma^2$ for a uniform distribution on $[a, b]$ is given by: $\sigma^2 = \frac{(b - a)^2}{12}$ Substituting $a = 7$ and $b = 16$: $\sigma^2 = \frac{(16 - 7)^2}{12} = \frac{9^2}{12} = \frac{81}{12} = 6.75$

3. Standard Deviation: The standard deviation $\sigma$ is the square root of the variance: $\sigma = \sqrt{6.75} \approx 2.598$

Thus, the standard deviation of $X$ is approximately $2.598$.

Would you like more details or have any questions?

Related Questions:

1. How do you find the variance of other continuous distributions?
2. What is the difference between standard deviation and variance?
3. How do you calculate the expected value of any random variable?
4. What is the significance of the uniform distribution in probability theory?
5. How does changing the interval $[a, b]$ affect the standard deviation?

Tip: For uniform distributions, the variance and standard deviation only depend on the length of the interval!